Simplify; express your answer in exponential form. Assume $r\neq 0, y\neq 0$. $\dfrac{{(ry^{5})^{-1}}}{{(r^{2}y^{-1})^{-4}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(ry^{5})^{-1} = (r)^{-1}(y^{5})^{-1}}$ On the left, we have ${r}$ to the exponent ${-1}$ . Now ${1 \times -1 = -1}$ , so ${(r)^{-1} = r^{-1}}$ Apply the ideas above to simplify the equation. $\dfrac{{(ry^{5})^{-1}}}{{(r^{2}y^{-1})^{-4}}} = \dfrac{{r^{-1}y^{-5}}}{{r^{-8}y^{4}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-1}y^{-5}}}{{r^{-8}y^{4}}} = \dfrac{{r^{-1}}}{{r^{-8}}} \cdot \dfrac{{y^{-5}}}{{y^{4}}} = r^{{-1} - {(-8)}} \cdot y^{{-5} - {4}} = r^{7}y^{-9}$